MySQL离开连接不使用AS关键字

Question

我喜欢将自解释名称用于关联选择，有时甚至必须避免重复，因此我使用AS关键字很多。但它给我带来了左连接的一些麻烦。

这有效：

$sql = "SELECT *,
        projects.id as projects_id
       FROM projects";

$sql .= " LEFT JOIN".
    " (SELECT 
        projectfiles.id as projectfiles_id,
        projectfiles.fileID as projectfiles_fileID,
        projectfiles.projectID as projectfiles_projectID
       FROM projectfiles
       ) AS projectfiles".
    " ON projects.id = projectfiles_projectID";

但是现在我最终得到了来自projects的无用数据，因为它还会选择我不需要的字段userID和name。它也是两次获取id。

所以我尝试将其改为;

$sql = "SELECT 
        projects.id as projects_id
       FROM projects";

ON线成为

" ON projects_id = projectfiles_projectID";

但是这给出了错误Unknown column projects_id

所以我试过

" ON projects.projects_id = projectfiles_projectID";

但仍然是同样的错误

然后我开始尝试，并尝试（作为测试）

$sql = "SELECT id,name,userID FROM projects";

$sql .= " LEFT JOIN".
    " (SELECT 
        projectfiles.id as projectfiles_id,
        projectfiles.fileID as projectfiles_fileID,
        projectfiles.projectID as projectfiles_projectID
       FROM projectfiles
       ) AS projectfiles".
    " ON projects.id = projectfiles_projectID";

令我惊讶的是，LEFT JOIN似乎根本没有接受任何东西。

代码：

$sql = "SELECT id,name,userID FROM projects";

$sql .= " LEFT JOIN".
    " (SELECT 
        projectfiles.id as projectfiles_id,
        projectfiles.fileID as projectfiles_fileID,
        projectfiles.projectID as projectfiles_projectID
       FROM projectfiles
       ) AS projectfiles".
    " ON projects.id = projectfiles_projectID";

$res = mysql_query($sql);
if(!$res) die(mysql_error());

if(mysql_num_rows($res) > 0)
{
    $rownum = 0;
    while($row = mysql_fetch_assoc($res))
    {
        print_r($row);
        echo "<br/><br/>";
    $rownum++;
    }
}

输出：

这很奇怪，因为projects中只有一行，projectfiles中只有一行有该项目ID ...我做错了什么？

Answer 1

使用简短形式的查询：

$sql = "SELECT projects.id,projects.name,projects.userID FROM projects LEFT JOIN
       projectfiles ON projects.id = projectfiles.projectID";

Answer 2

SELECT p.*, pf.id, pf.fileId
FROM projects p LEFT JOIN projectfiles pf 
on p.id = pf.projectID

您可以使用“as”来执行操作。不需要子选择。

Answer 3

仅从projectfiles表中选择：

$sql = "SELECT projectfiles.*,
        projects.id as projects_id
       FROM projects";

// rest of the code is the same

<强>更新

$sql = "SELECT projectfiles.* FROM projects";
// rest of the code is the same

Answer 4

$sql = "SELECT p.id,p.name,p.userID FROM projects p";

$sql .= " LEFT JOIN".
    " projectfiles pf ".
    " ON p.id = p.projectID";

Answer 5

$sql = "SELECT prj.id as prjId,
    prj.name as prjName,
    prj.userID as prjUid,
    pf.id as pfId, 
    pf.fileID as pfFileId, 
    pf.projectID as pfProjecId 
FROM projects as prj 
LEFT JOIN projectfiles AS pf 
ON prj.id = pf.projectID";